The Need for the Second Law of Thermodynamics
The first law of thermodynamics states that when one form of energy is transformed
into another, the total energy is conserved, but does not indicate any other restriction
on this process.
Advanced Physical Chemistry
E NTROPY
C HAPTER 5
AND THE S ECOND AND T HIRD L AWS
OF T HERMODYNAMICS
Yet, we know that processes and chemical reactions proceed spontaneously in one
direction, BUT NOT in the opposite direction.
The first law of thermodynamics does not tell us the direction in which a process can
occur spontaneously.
The laws of classical mechanics and quantum mechanics also do not tell us the
direction in which a process can be spontaneous.
Professor Angelo R. Rossi
It is useful to be able to predict whether a physical change, or a chemical reaction, will
proceed spontaneously in the forward or backward direction, and so the second law of
thermodynamics is very important.
http://homepages.uconn.edu/rossi
Department of Chemistry, Room CHMT214
The University of Connecticut
Fall Semester 2013
angelo.rossi@uconn.edu
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Introducing Entropy to Identify Spontaneous Processes
What is Entropy?
The first law shows that when an ideal gas is heated reversibly
Entropy is that thermodynamic property that describes the disorderly
disposal of the total energy of an isolated system.
dU = CV dT = dqrev + dw = dqrev − P dV = dqrev
and
dqrev
nRT
−
dV
V
It is a state function which is a measure of disorder in a system and a
signpost for spontaneous change.
nRT
= CV dT +
dV
V
The Second Law of Thermodynamics states that the entropy of an
isolated system increases in the course of spontaneous change.
Taking the derivatives of dT and dV :
∂CV
∂V
=0
T
but
∂( nRT
)
V
∂T
!
=
V
nR
V
∆Stotal > 0
where ∆Stotal is the total entropy of the system plus surroundings.
Since dq is an inexact differential, use an integrating ( T1 ) factor to convert an inexact differential to an exact
differential:
CV
nR
dqrev
=
dT +
dV
T
T
V
!
!
∂( nR
∂( CTV )
)
V
= 0 but
=0
∂V
∂T
T
2
V
and ∆S = dqTrev is a state function and new thermodynamic variable which serves as marker for
spontaneity.
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Thermodynamic Definition of Entropy
Entropy Change for an Isothermal Process
Calculate the entropy change of a sample of a perfect gas when it expands isothermally from V1 to V2 .
The thermodynamic definition of entropy concentrates on the dispersion of energy as
the result of a physical or chemical change that is transferred in the form of heat.
dS =
1.
Find the heat absorbed for a reversible path between the initial and final states independent of the
manner in which the process takes place.
dqrev
T
∆U = q + w
qrev = −w (for a reversible change)
Entropy is a state function and is measured between an initial state (i) and final state
(f):
Z f
dqrev
∆S =
T
i
2.
Isothermal process, T is constant, so T can be taken outside the integral.
∆S =
To calculate the entropy between two states of a system
1.
2.
∆U = 0
Find a reversible path between the states
Integrate the heat supplied along the path divided by the temperature
at which the heat is supplied.
1
T
Z
f
dqrev =
i
qrev = −w = nRT ln
∆S = nR ln
3.
qrev
T
Vf
Vi
Vf
Vi
For example, when the volume of 1.00 mole of perfect gas is doubled at a constant temperature:
∆S = (1.00 mol)(8.3145 J K−1 mol−1 ) × ln 2 = +5.76 J K−1
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Change in Entropy of the Surroundings
Entropy as a State Function
Consider the surroundings as reservoir of constant volume to which an
infinitesimal transfer of heat is made:
Entropy is a state function and independent of path
I
dqrev
=0
T
H
where denotes integration around a closed path.
∆Usur = dqsur
Since ∆Usur is an exact differential, it is independent of how the change
is brought about. It can be either reversible or irreversible.
Since dqsur = ∆Usur ,
dqsur
∆Ssur =
Tsur
For a measurable change in temperature,
∆Ssur =
qsur
Tsur
The Path of a Thermodynamic Cycle
The above can be performed either reversibly or irreversibly.
For an adiabatic process, qsur = 0 and ∆Ssur = 0.
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The entropy is the same at the initial and final states and is independent of the path
taken between them.
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Entropy as a State Function: The Carnot Cycle
A General Cycle
The Carnot cycle consists of taking a sample of an ideal gas
through four reversible stages:
1.
2.
3.
4.
Any reversible cycle can be approximated by a collection of
Carnot cycles, and the cyclic integral around an arbitrary path
is the sum of the integrals around each Carnot cycle.
Reversible isothermal expansion of the gas from state A to
state B where qh is the heat supplied from the hot source.
The entropy change is given by ∆S1 = Tqh .
h
Reversible adiabatic expansion of the gas from state B to
state C. No heat leaves the system so the temperature
falls from Th to Tc , and ∆S2 = 0.
Reversible isothermal compression of the gas from state
C to state D, and qc is negative, and ∆S3 = Tqc
c
Reversible adiabatic compression of the gas from state
D to state, and the temperature rises from Tc to Th , and
∆S4 = 0.
All the entropy integrals cancel except for those along the
perimeter of the overall cycle:
X qrev
=
T
all
dS = ∆S1 + ∆S2 + ∆S3 + ∆S4 =
But
qrev
=0
T
In the limit of infinitismal cycles, the non-cancelling edges of
the Carnot cycle match the overall cycle exactly, and the sum
becomes the integral:
The total entropy change around the cycle is
I
X
perimeter
I
qc
qh
+
Th
Tc
dS = 0
A General Cycle
qh
Th
=
qc
Tc
for an ideal gas. Thus
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I
A Typical Carnot Cycle
dS = 0
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The Clausius Inequality
The Clausius Inequality
It is important to verify that entropy is an indicator of spontaneous change.
For any change then,
•
10
dq
T
If the system is isolated from the surroundings, then dq = 0, and the Clausius
inequality implies that dS ≥ 0.
Consider a system in thermal and mechanical contact with its surroundings at the
same temperature:
•
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dS ≥
A change of state is accompanied by a change in entropy of the system (dS) and
the surroundings (dSsur ).
Because the process might be irreversible, the total entropy will increase when this
process occurs in the system:
This tells us that, in an isolated system, entropy cannot decrease when a spontaneous
change takes place.
Thus, entropy is an indicator of spontaneous change.
dS + dSsur ≥ 0
For any spontaneous change, dStot ≥ 0, and for a reversible process, dStot = 0.
then
dS ≥ −dSsur
The entropy change for the surroundings is given by
dSsur = −
dq
T
because heat from the surroundings enters the system.
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The Second Law of Thermodynamics
Entropy Changes in Reversible and Irreversible Processes
There are two parts to the Second Law of Thermodynamics:
1.
2.
(a) Irreversible expansion of
an ideal gas at 298 K into
twice the initial volume with
no heat or work.
There is a state function called entropy (S) that can be calculated
from
dqrev
dS =
T
The change in entropy for any process is given by
dS ≥
(b) Reversible isothermal
expansion of an ideal gas
to twice the initial volume
at 298 K. In this case the
piston is surrounded by a
reservoir
dq
T
where (>) applies to a spontaneous irreversible process, and (=)
applies to a reversible process.
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∆S for a System and Surroundings in a Reversible Process
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(a) Irreversible expansion of
an ideal gas at 298 K into
twice the initial volume with
no heat or work.
What is the change in entropy of the gas?
qrev = −wrev = nRT ln
∆S = nR ln
(b)
How much work is done on the gas?
wrev = −nRT ln
V2
V1
V2
V1
= (0.5 mol) (8.3145 J K
Vf
Vi
−1
mol
−1
mol
= (0.5 mol)(8.3145 J K
(c)
What is qsurr ?
Since this is an ideal gas, there is no change in internal energy: ∆U = q + w = 0.
(d)
What is the change in entropy of the surroundings?
Since heat flows out of the surroundings, it has a decrease in entropy:
qsys = +859 J
∆Ssurr =
(e)
-859 J
298.15 K
−1
) ln 2 = 2.88 J K
−1
)(298.15 K) ln 2 = - 859 J
−1
(b) Reversible isothermal
expansion of an ideal gas
to twice the initial volume
at 298 K. In this case the
piston is surrounded by a
reservoir
qsurr = -859 J
= -2.88 J K
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Entropy Changes in Irreversible Processes
One half mole of an ideal gas expands isothermally and reversibly at 298.15 K from a volume of 10 L to a volume of 20 L.
(a)
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−1
What is the change in entropy of the system plus surroundings?
Since the entropy of the gas increases, and the entropy of the surroundings decreases, there is no change in entropy for the system and
the surroundings.
The gas and the surroundings can be considered to be an isolated system; the process is reversible, and so ∆S = 0
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∆S for a System and Surroundings in an Irreversible Process
Entropy Changes in Reversible Processes - Phase Transitions
Now consider the same expansion in the preceding example occurs irreversibly by
simply opening a stopcock and allowing the gas to rush into an evacuated container of
10 L volume.
Consider the vaporization of a pure liquid at the equilibrium vapor pressure, P
liquid(T, P ) → vapor(T, P )
(a) What is the change in entropy of the gas?
The change in entropy is the same as for the reversible process because entropy is
a state function.
(b) How much work is done on the gas?
No work is done in the expansion.
(c) What is qsurr ?
No heat is exchanged with the surroundings.
(d) What is the change in entropy of the surroundings?
The entropy of the surroundings does not change.
(e) What is the change in entropy of the system plus surroundings?
The entropy of the system plus surroundings increases by 2.88 J K−1 .
Since this is an irreversible process, we expect the entropy to increase.
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Since T is constant
qrev
∆Hvap
=
T
T
A similar equation can be used to calculate the entropy of sublimation, the entropy of
melting, and the entropy change for a transition between two forms of a solid.
∆S =
The molar entropy of a vapor is always greater than that of the liquid with which it is in
equilibrium.
The molar entropy of a liquid is always greater than that of a solid at its melting point.
Entropy is a measure of the disorder in a system, and the molecules in a vapor are
more disordered than those in the liquid, and the molecules of the liquid are more
disordered than those in a solid.
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Application of the Second Law to the Vaporization of a Liquid
Heating and Cooling of Substances
The liquid and vapor (the system) and the heat reservoir (the surroundings) form an
isolated overall system where the total entropy change is given by:
Heating or cooling of substances can be carried out reversibly.
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The entropy change when a substance is heated or cooled reversibly can be
calculated using:
∆S = ∆Ssys + ∆Ssurr
The heat gained by the system is equal to that lost by the surroundings.
dS =
The entropy change for the surroundings is the negative of the entropy change for the
system, if the vaporization is carried out reversibly.
CP
qrev
=
dT (constant pressure process)
T
T
or
qrev
CV
=
dT (constant volume process)
T
T
If the heat capacity is independent of temperature, and the temperature is changed
from T1 to T2 , then for a constant volume process,
dS =
For both the system and surroundings, ∆S = 0.
∆S =
Z
T2
T1
T2
CV
dT = CV ln
T
T1
A similar expression is obtained for a constant pressure process.
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Heating a Substance without a Phase Change
Change of Entropy with Temperature
The entropy of a system at temperature Tf can be calculated from a
knowledge of the entropy at Ti :
Z Tf
dqrev
S(Tf ) = S(Ti ) +
T
Ti
For a constant pressure process, dqrev = CP dT , then
Z Tf
CP
S(Tf ) = S(Ti ) +
dT
T
Ti
When a substance is heated from T1 to T2 without a phase change, the
increase in entropy is given by the indicated area.
If the volume is constant, CV is used, and if the pressure is constant, CP
is used.
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Variation of Entropy with Temperature
The Third Law of Thermodynamics
Oxygen is heated from 300 K to 500 K at a constant pressure of 1 bar.
What is the increase in molar entropy?
The molar heat capacity (CP,m ) of O2 has the following dependence on
temperature:
CP,m = α + βT + γT 2
The entropy change for certain isothermal chemical reactions approachs zero as the
temperature is reduced.
The entropy cannot be determined calorimetrically using ∆S =
reactions do not occur reversibly.
∆S = 15.41 J K
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mol
because chemical
It is possible to discuss the measurement of the change in entropy for a phase
change of a single substance such as the transformation from the rhombic form of
sulfur to the monoclinic form of sulfur:
α = 25.460 J K−1 mol−1
β = 1.519 x 10−2 J K−2 mol−1
γ = -0.715 x 10−5 J K−3 mol−1
Z T2
Z T2 CP,m
α
∆S =
dT =
+ β + γT dT
T
T
T1
T1
γ 2
T2
+ β (T2 − T1 ) +
T2 − T12
∆S = α ln
T1
2
−1
qrev
T
S (rhombic) ⇌ S (monoclinic)
The change in entropy for a phase change approaches zero as the temperature is
reduced to absolute zero.
Z T
CP,m
dT
ST,m − S0,m =
T
0
Rhombic sulfur is the stable form below the transition temperature of 368.5 K.
−1
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The Third Law of Thermodynamics
The Third Law of Thermodynamics
Monoclinic sulfur can be supercooled below 368.5 K and its heat capacity measured
in the vicinity of T = 0.
Molar entropies of the monoclinic
and rhombic forms of sulfur from
absolute zero to the transition
temperature (368.5 K)
The Debye Extrapolation is a way to obtain the entropy of a substance as
temperature approaches T ≈ 0. In this situation, it is assumed that the CP ∝ T 3 :
CP = aT 3 (curve near T = 0 is fit to data)
The entropy change for the phase change
S (rhombic) ⇌ S (monoclinic)
It can be seen in the above figure that as T → 0, ∆Stransition → 0.
at the transition temperature can be experimentally determined to be
1.09 J K−1 mol−1 based on the assumption that S0,m = 0 for both the rhombic and
monoclinic forms of sulfur.
This behavior was found for many other substances by W. Nernst which led him to the
conclusion:
lim ∆Sreaction = 0
This is in agreement with the entropy change at 368.5 K calculated from the
difference in enthalpy between the two forms (∆H = 401 J mol−1 ):
as the temperature approaches 0 K, the ∆Sreaction approaches zero for all reactions.
∆Stransition
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T →0
Third Law of Thermodynamics: the entropy of each pure element of substance in a
perfect crystalline form is zero at absolute zero.
401 J mol−1
= 1.09 J K−1 mol−1
=
368.5 K
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Third-Law Entropies
Fundamental Equations of Thermodynamics
Entropies which are given on the basis of S(0) = 0 are called
Third-Law Entropies.
Although the entropy provides a criterion of whether a change in an
isolated system is spontaneous, it does not provide a convenient
criterion at constant T and V or constant T and P which are usual
conditions in the laboratory.
The standard entropy has the symbol S ◦ and are tabulated at 298.15
K and a pressure of 1 bar.
Two additional thermodynamic functions are needed to make
calculations at constant T and V or T and P more convenient than is
possible using the entropy.
The standard reaction entropy is defined as the difference in
entropies between pure products and pure reactants in their standard
states at the specified temperature:
X
X
◦
◦
−
νSm
∆Sreaction =
νSm
P roducts
These two new thermodynamic properties are the Helmholtz energy
(A) and the Gibbs energy (G):
Reactants
At constant T and V , spontaneous processes occur with a decrease in
A.
where ν stands for the appropriate coefficients in the balanced chemical
reaction.
At constant T and P , spontaneous processes occur with a decrease in
G.
Standard reaction enthalpies are usually positive when a gas is formed
in a reaction, and negative when gas is consumed in a chemical
reaction.
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Fundamental Equation for the Internal Energy
Properties of the Internal Energy
The First Law of thermodynamics is written
The equation for internal energy
dU = T dS − P dV
dU = dq + dw
and the second law is given as
suggests that U is a function of S and V
dq
T
where the inequality applies for dq in a irreversible process, and the equality relates to
a reversible process.
For a reversible change in a closed system with constant composition in the absence
of additional, non-expansion work:
dS ≥
U = U (S, V )
thus
dU =
The combined first and second law is
The above equation shows that U is a function of S and V and are called natural
variables of U .
The natural variables for the internal energy are all extensive.
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Maxwell Relations
dS +
V
∂U
∂V
dV
S
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New Thermodynamic Potentials
For the infinitesimal change of a function f(x,y) written as
The internal energy (U ) and enthalpy (H) are referred to a thermodynamic potentials.
To define new thermodynamic potentials, one can use the method of Legendre transforms.
df = g(x, y)dx + h(x, y)dy
•
The mathematical criterion for df being an exact differential is
∂g(x, y)
∂y
=
x
∂h(x, y)
∂x
•
A Legendre transform is a linear change of variables that starts with a mathematical function and defines
a new function by subtracting one or more conjugate variables.
A Legendre transform was already used to define the enthalpy
H = U + PV
y
To provide a more complete treatment of enthalpy, the total differential is given by
Because the fundamental Equation for the internal energy
dH = dU + P dV + V dP
dU = T dS − P dV
is an expression for an exact differential, one can write
One can write
Note that the derivatives of extensive properties with respect to extensive properties
are intensive properties.
dU = T dS − P dV
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∂U
∂S
Comparing these two equations, one obtains
∂U
∂U
=T
= −P
∂S V
∂V S
dwrev = −P dV
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∂U
∂S
=T
V
∂T
∂V
S
=−
∂U
∂V
∂P
∂S
S
Substituting for U yields the fundamental equation for the enthalpy of a system involving pressure-volume
work
dH = T dS + V dP
= −P
and
∂H
∂S
P
=T
∂H
∂P
=V
S
V
The above equation is an example of a Maxwell Relation.
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Two New Thermodynamic Potentials:
Helmholtz and Gibbs Energies
The Helmholtz Energy
The internal energy and enthalpy do not provide very useful criteria for
spontaneous change, because the entropy has to be held constant.
The total differential of the Helmholtz energy is given by
This problem can be fixed by using a Legendre transform in which the
product T S is subtracted from either U or H.
dA = dU − T dS − SdT
Substituting for dU , the fundamental equation for the Helmoltz energy becomes
The Helmholtz energy, A is given by
dA = SdT − P dV
A = U − TS
and the natural varaibles for A are T and V .
∂A
∂A
=S
−
=P
−
∂T V
∂V T
while the Gibbs energy, G is given by
G = U + PV − TS = H − TS
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(Helmholtz Energy)
A = U − TS
New Maxwell relations are also provided by the Helmholtz potential.
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The Helmholtz Energy
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The Gibbs Energy
The total differential for the Gibbs energy is
The Helmholtz energy, A provides a criterion for spontaneity at constant volume and temperature.
dG = dU + P dV + V dP − T dS − SdT
dAT,V ≤ 0
Substituting for dU yields the fundamental equation for the Gibbs energy
T and V are constant.
The criterion for equilibrium when neither the forward nor the reverse reaction change is
dG = −SdT + V dP
dAT,V = 0
Given that G = G(T, P ) allows
dG =
The change in Helmholtz energy is equal to the maximum work accompanying a process:
∂G
∂T
P
dT +
∂G
∂P
dP
T
Comparing the above equation with the fundamental equation for the Gibbs energy yields
dA = dwmax
When a macroscopic isothermal change takes place in the system, the following equations are obtained:
P
= −S
∂G
∂P
=V
T
which shows how the Gibbs energy varies with temperature and pressure.
wmax = ∆A
At constant temperature and pressure, chemical reactions which are spontaneous have an accompanying
decrease in Gibbs energy: dGT,P ≤ 0.
and
∆A = ∆U − T ∆S
The Helmholtz energy is less useful in chemistry than the Gibbs energy because reactions are more often
carried out at constant pressure rather than constant volume.
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∂G
∂T
The criterion for equilibrium when neither the forward nor the reverse reaction change is dGT,P = 0
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Properties of the Gibbs Energy
Properties of the Gibbs Energy
The partial derivatives obtained on the previous slide provide an analysis of the Gibbs
energy.
∂G
∂G
= −S
=V
∂T P
∂P T
Because S is positive, it follows that G decreases when the temperature is raised at
constant composition.
G decreases sharply when the entropy change for a system is large. But, the Gibbs
energy of a gaseous substance is more sensitive to temperature than the liquid and
solid phases.
Because V is positive, G always increases when the pressure of the system increases
at constant temperature and composition.
Variation of the Gibbs energy of a system with temperature at constant pressure and pressure at constant temperature. The slope of the Gibbs energy with
respect to pressure is equal to V .
G is more sensitive to pressure changes for the gaseous phase than for its liquid or
gaseous phases.
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Variation of the Gibbs energy with temperature is determined by the entropy.
Because the entropy of a gas > entropy
of a liquid > entropy of a solid, the Gibbs
energy changes the most steeply for a
gas.
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The Gibbs Energy as a Predictor of Spontaneous Change
Standard Molar Gibbs Energies
G can be used to determine the direction of spontaneous change in a chemical
transformation.
For macroscopic changes at constant T and P , the condition for spontaneity is
∆G < 0 where
∆G = ∆H − T ∆S
Standard entropies and enthalpies can be combined to obtain the standard Gibbs
energy of a reaction:
◦
◦
− T ∆Sreaction
∆G◦reaction = ∆Hreaction
where
Thus, there are two contributions to ∆G that determine if an isothermal chemical
transformation is spontaneous
1.
2.
X
◦
∆Hreaction
=
P roducts
the energetic contribution, ∆H.
the entropic contribution, T ∆S.
◦
=
∆Sreaction
X
ν∆Hf◦ −
P roducts
The following conclusions can be drawn from the above equation
•
•
•
•
38
◦
−
νSm
X
ν∆Hf◦
Reactants
X
◦
νSm
Reactants
It is also possible to define the standard Gibbs energy of formation as the standard
reaction Gibbs energy for the formation of a compound from its elements in their
standard states:
X
X
ν∆G◦f
ν∆G◦f −
∆G◦reaction =
The entropic contribution for ∆G is greater for higher temperatures.
A chemical transformation is always spontaneous if ∆H < 0 and ∆S > 0.
A chemical transformation is never spontaneous if ∆H > 0 and ∆S < 0.
For all other cases, the relative magnitudes of ∆H and T ∆S determine if the
chemical transformation is spontaneous.
P roducts
Reactants
where the standard free energies of the elements in their reference states are zero.
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Effect of Temperature on the Gibbs Energy
Effect of Pressure on the Gibbs Energy
The change in Gibbs energy can be expressed in terms of enthalpy (H) by making the following substitution:
∂G
∂T
P
= −S =
Liquids and Solids
G−H
T
Keeping the temperature fixed and determining the Gibbs energy at a new pressure
yields
✿0
✘✘
SdT
dG = V dP − ✘
This equation can be rearranged to yield
∂ G
∂T T
P
=−
and
H
T2
G(Pf ) = G(Pi ) +
Z
Pf
V dP
Pi
Liquids and solids undergo small volume changes as the pressure changes, so that V
can be treated as a constant and taken out of the integral:
The above expression is called the Gibbs-Helmholtz equation which indicates that it is possible to
varies with temperature if the enthalpy of the system is known.
determine how G
T
The Gibbs energy between the initial and final states is given as ∆G = Gf − Gi , and the above equation
can be applied to both states:
∂ ∆G
∆H
=− 2
∂T T
T
P
Under normal laboratory conditions, Vm ∆P is small and may be neglected.
This equation indicates how the change in Gibbs energy varies with temperature, if the change in enthalpy
of the system undergoing the change is also known.
However, if the pressure changes are extremely large, then the effect of pressure on
solids and liquids must be taken into account.
Gm (Pf ) = Gm (Pi ) + Vm (Pf − Pi ) = Gm (Pi ) + Vm ∆P
The Gibbs-Helmholtz equation is most useful when applied to changes of physical state and chemical
reactions at constant pressure.
Fall 2013
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Last Updated: September 20, 2013 at 4:49pm
Effect of Pressure on the Gibbs Energy
Fall 2013
Last Updated: September 20, 2013 at 4:49pm
42
Effect of Pressure on the Gibbs Energy
Gases
For a perfect gas, V =
nRT
P
can be substituted in the equation below:
Gm (Pf ) = Gm (Pi ) +
Z
Pf
V dP = Gm (Pi ) + nRT
Pi
and
Gm (Pf ) = Gm (Pi ) + nRT ln
Z
Pf
Pi
dP
P
Pf
Pi
If the initial pressure is set to the standard pressure at 1 bar (Pi = P ◦ ), the Gibbs
energy of a perfect gas at a pressure, P , is related to its standard value by
Gm (P ) = Gm (P ◦ ) + nRT ln
The variation of the Gibbs energy with the
pressure is determined by the volume of
the sample. Because the volume of a
gaseous phase of a substance > than
the same amount of a liquid phase and
> than the solid phase, the Gibbs energy
changes most steeply for the gas then
the liguid, followed by the solid phase.
P
P◦
In going from an idealized system to a real system, it is important to preserve the form
of the equations that have been derived for the idealized system.
This allows deviations from idealized behavior to be presented most simply.
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Fall 2013
The molar Gibbs energy potential is proportional to ln P , and the standard state
is reached at P ◦ .
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